📚 A-Level Further Mathematics Unit 3 June 2019 Mark Scheme: Key Concepts | A-Level 进阶数学单元3 2019年6月评分标准知识点精讲
This article provides a detailed revision of the key topics assessed in a typical A-Level Further Mathematics Unit 3 paper, using the June 2019 mark scheme as a guide. By examining the concepts behind the mark scheme entries—such as complex loci, matrix transformations, polar area, hyperbolic identities, and reduction formulae—you can deepen your understanding of the required techniques. Each section presents the theory, common pitfalls, and examples styled after real exam questions, helping you master the content for top marks.
本文以2019年6月A-Level进阶数学单元3评分标准为线索，系统梳理考试核心知识点。通过深入剖析复数轨迹、矩阵变换、极坐标面积、双曲恒等式和递推积分等评分条目背后的原理，结合典型真题示例与常见错误解析，帮助学生扎实掌握每个技巧，确保在实际考试中能准确应用并拿到全部分数。
1. Complex Numbers – Loci and Regions | 复数 – 轨迹与区域
Complex loci often appear in the form |z – a| = r or arg(z – a) = θ. The June 2019 mark scheme rewards clear geometric interpretation: the first represents a circle centred at a with radius r, while the second is a half-line from a at angle θ to the positive real axis, excluding the point a itself. When shading a region given by an inequality such as |z – 3 – 4i| < 5, you must show a dashed boundary to indicate strict inequality and shade the appropriate interior or exterior. Common mistakes include forgetting to convert cartesian coordinates (3,4) correctly into modulus form and failing to indicate the excluded endpoint on a half-line.
复数轨迹常以 |z – a| = r 或 arg(z – a) = θ 的形式出现。2019年6月评分标准强调清晰的几何解释：前者表示以 a 为圆心、r 为半径的圆；后者表示从 a 出发、与正实轴夹角为 θ 的射线（不包含点 a 本身）。若区域由不等式如 |z – 3 – 4i| < 5 定义，需用虚线表示边界（严格不等），并正确填充内部或外部。常见错误包括：将笛卡尔坐标 (3,4) 转换为模时出错，或在射线上未标明排除的端点。
Example: Sketch the locus of z such that |z + 2i| = |z – 4| and find its Cartesian equation.
示例： 绘制满足 |z + 2i| = |z – 4| 的轨迹并求其笛卡尔方程。
Let z = x + iy, then equate moduli: √(x² + (y+2)²) = √((x–4)² + y²). Squaring and simplifying yields 8x + 4y – 12 = 0, i.e., 2x + y = 3, which is a straight line. The mark scheme requires this algebraic derivation, not just the geometric perpendicular bisector statement.
设 z = x + iy，模相等得 √(x² + (y+2)²) = √((x–4)² + y²)。平方化简得 8x + 4y – 12 = 0，即 2x + y = 3，为一直线。评分标准要求展示这种代数推导，而非仅陈述几何上的中垂线结论。
2. De Moivre’s Theorem and Trigonometric Identities | 棣莫弗定理与三角恒等式
De Moivre’s theorem states that (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ for integer n. In the June 2019 paper, you might be asked to express cos 5θ in terms of powers of cos θ by expanding (c + is)⁵ and equating real parts. The mark scheme requires a systematic binomial expansion: cos 5θ = 16 cos⁵θ – 20 cos³θ + 5 cos θ. Students often lose marks by misapplying i powers or by forgetting to use sin²θ = 1 – cos²θ to replace remaining sin² terms. Similarly, to find tan 5θ, you use the sine expansion divided by the cosine expansion.
棣莫弗定理表明 (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ，n 为整数。2019年6月试题中可能要求将 cos 5θ 表示为 cos θ 的多项式，通过展开 (c + is)⁵ 并取实部。评分标准重视系统的二项式展开，最终结果 cos 5θ = 16 cos⁵θ – 20 cos³θ + 5 cos θ。学生常因 i 的幂次处理失误或忘记用 sin²θ = 1 – cos²θ 消去 sin² 项而失分。类似地，求 tan 5θ 时需用正弦展开式除以余弦展开式。
Example: Using De Moivre, prove that cos 4θ = 8 cos⁴θ – 8 cos²θ + 1.
示例： 利用棣莫弗定理证明 cos 4θ = 8 cos⁴θ – 8 cos²θ + 1。
Expand (cos θ + i sin θ)⁴ = cos 4θ + i sin 4θ. Binomial expansion: cos⁴θ + 4i cos³θ sin θ – 6 cos²θ sin²θ – 4i cos θ sin³θ + sin⁴θ. Real part: cos⁴θ – 6 cos²θ sin²θ + sin⁴θ. Replace sin²θ = 1 – cos²θ, then sin⁴θ = (1 – cos²θ)². After simplification you obtain 8 cos⁴θ – 8 cos²θ + 1, which is the required identity.
展开 (cos θ + i sin θ)⁴ = cos 4θ + i sin 4θ。二项式展开实部为 cos⁴θ – 6 cos²θ sin²θ + sin⁴θ。代入 sin²θ = 1 – cos²θ，并将 sin⁴θ 写成 (1 – cos²θ)²，化简后得到 8 cos⁴θ – 8 cos²θ + 1。
3. Matrix Transformations and Eigenvalues | 矩阵变换与特征值
In Further Mathematics Unit 3, matrix questions often ask for eigenvalues and eigenvectors of a 2×2 or 3×3 matrix, and then to interpret the transformation geometrically. The June 2019 mark scheme awards marks for correctly solving det(A – λI) = 0 and for finding corresponding eigenvectors by solving (A – λI)x = 0. Typical pitfalls include algebraic mistakes in the determinant expansion, not expressing eigenvectors in a simplified integer direction, or forgetting to state that a line of invariant points corresponds to an eigenvalue of 1.
进阶数学单元3的矩阵题常要求计算 2×2 或 3×3 矩阵的特征值与特征向量，并进行几何解释。2019年评分标准对正确求解 det(A – λI) = 0 及通过 (A – λI)x = 0 求特征向量给予分值。常见错误有：行列式展开时代数出错、特征向量未化为最简整数比，或未指出特征值 1 对应不变点线。
For a matrix M representing a linear transformation, invariant lines are found by setting y = mx + c and solving M(x, y) = (x‘, y‘) with y‘ = mx‘ + c. Often the mark scheme accepts the method of solving for m and c from simultaneous equations.
对于变换矩阵 M，不变直线可通过设 y = mx + c 并使 M(x, y) 映射后的点满足 y‘ = mx‘ + c 来求解。评分标准认可从联立方程组中解出 m 和 c 的方法。
Example table for eigenvalues and invariant lines:
特征值与不变直线示例表格：
| Matrix | Eigenvalues | Invariant lines |
|---|---|---|
| (4 2; -1 1) | λ = 2, 3 | y = –x, y = –½x |
4. Polar Coordinates – Area and Tangents | 极坐标 – 面积与切线
The area enclosed by a polar curve r = f(θ) from θ = α to β is (1/2) ∫ r² dθ. The June 2019 mark scheme frequently tested finding the area of a loop or region bounded by two polar curves. A common trick is to use symmetry to double the integral from 0 to the half-loop angle. When finding points where a tangent is parallel to the initial line, set dy/dθ = 0 using y = r sin θ; for perpendicular, set dx/dθ = 0 where x = r cos θ. Candidates must correctly differentiate products such as r sin θ.
极坐标曲线 r = f(θ) 在 α 到 β 之间所围面积公式为 (1/2) ∫ r² dθ。2019年6月试题经常考查求曲线环圈或两曲线围成的区域面积。常用技巧是利用对称性，将积分从 0 到半圈角度加倍计算。求平行于极轴的切线时，利用 y = r sin θ，令 dy/dθ = 0；求垂直切线则令 x = r cos θ 的导数 dx/dθ = 0。考生必须正确求导诸如 r sin θ 这样的乘积。
Example: Find the area of the inner loop of r = 1 + 2 cos θ.
示例： 求 r = 1 + 2 cos θ 内环的面积。
The inner loop exists between θ = 2π/3 and 4π/3 where r ≤ 0. Area = (1/2) ∫ (1 + 2 cos θ)² dθ from 2π/3 to 4π/3. Expand and integrate term by term, using cos²θ = (1+cos 2θ)/2. The final exact area is π – (3√3)/2. Mark schemes expect accurate limits and correct integration of trigonometric squares.
内环出现在 r ≤ 0 时，即 θ 从 2π/3 到 4π/3。面积 = (1/2) ∫ (1 + 2 cos θ)² dθ 在该区间积分。展开并逐项积分，使用 cos²θ = (1+cos 2θ)/2，精确结果为 π – (3√3)/2。评分标准要求积分限准确且三角平方积分无误。
5. Hyperbolic Functions – Identities and Calculus | 双曲函数 – 恒等式与微积分
Hyperbolic functions are defined as sinh x = (eˣ – e⁻ˣ)/2, cosh x = (eˣ + e⁻ˣ)/2, and tanh x = sinh x / cosh x. The mark scheme routinely tests the identity cosh²x – sinh²x = 1 and analogous double-angle formulas: cosh 2x = cosh²x + sinh²x = 2 cosh²x – 1 = 1 + 2 sinh²x. Differentiation is straightforward: d/dx (sinh x) = cosh x, d/dx (cosh x) = sinh x. Integration of hyperbolic functions often uses the same rules as trigonometric ones but without sign changes. For instance, ∫ cosh 3x dx = (1/3) sinh 3x + C.
双曲函数定义为 sinh x = (eˣ – e⁻ˣ)/2，cosh x = (eˣ + e⁻ˣ)/2，tanh x = sinh x / cosh x。评分标准常考恒等式 cosh²x – sinh²x = 1 及其倍角公式：cosh 2x = cosh²x + sinh²x = 2 cosh²x – 1 = 1 + 2 sinh²x。求导很容易：d/dx (sinh x) = cosh x，d/dx (cosh x) = sinh x。双曲函数的积分法则与三角类似但无符号变化，例如 ∫ cosh 3x dx = (1/3) sinh 3x + C。
Example: Solve the equation 5 sinh x + 3 cosh x = 4, giving the answer in logarithmic form.
示例： 求解方程 5 sinh x + 3 cosh x = 4，用对数形式表示答案。
Express in exponentials: 5(eˣ – e⁻ˣ)/2 + 3(eˣ + e⁻ˣ)/2 = 4. Multiply by 2: (5eˣ – 5e⁻ˣ + 3eˣ + 3e⁻ˣ) = 8 → 8eˣ – 2e⁻ˣ = 8. Multiply by eˣ: 8e²ˣ – 2 = 8eˣ → 4e²ˣ – 4eˣ – 1 = 0. Treat as quadratic in eˣ: eˣ = (4 ± √(16 + 16))/8 = (4 ± √32)/8 = (4 ± 4√2)/8 = (1 ± √2)/2. Since eˣ > 0, take eˣ = (1 + √2)/2, thus x = ln((1+√2)/2). The mark scheme gives credit for converting to exponentials and solving the quadratic.
用指数表示：5(eˣ – e⁻ˣ)/2 + 3(eˣ + e⁻ˣ)/2 = 4。乘2整理得 8eˣ – 2e⁻ˣ = 8。两边乘 eˣ 得 8e²ˣ – 2 = 8eˣ → 4e²ˣ – 4eˣ – 1 = 0。解 eˣ 的二次方程，取正值得 eˣ = (1+√2)/2，因此 x = ln((1+√2)/2)。评分标准认可转换为指数并正确解二次方程的过程。
6. Further Calculus – Reduction Formulae | 进阶微积分 – 递推公式
Reduction formulae involve establishing a recurrence relation for integrals of the form Iₙ = ∫ f(x,n) dx, often using integration by parts. A typical June 2019 question might ask for Iₙ = ∫₀^{π/2} sinⁿx dx, deriving Iₙ = ((n-1)/n) Iₙ₋₂. The mark scheme heavily weights the correct application of integration by parts with appropriate choices for u and dv, and careful manipulation of limits. After obtaining the reduction formula, candidates must then use it to evaluate specific integrals like I₄ or I₅.
递推公式涉及通过分部积分建立形如 Iₙ = ∫ f(x,n) dx 的递推关系。2019年6月试题可能考查 Iₙ = ∫₀^{π/2} sinⁿx dx，推导出 Iₙ = ((n-1)/n) Iₙ₋₂。评分标准侧重正确选择 u 和 dv 进行分部积分，并细致处理积分限。得到递推式后，再将其应用于计算具体积分如 I₄ 或 I₅。
Example: Given Iₙ = ∫₀¹ xⁿ eˣ dx, show that Iₙ = e – n Iₙ₋₁ for n ≥ 1, and hence find I₃.
示例： 已知 Iₙ = ∫₀¹ xⁿ eˣ dx，证明 Iₙ = e – n Iₙ₋₁（n ≥ 1），并由此求 I₃。
Integration by parts: let u = xⁿ, dv = eˣ dx → du = n xⁿ⁻¹ dx, v = eˣ. Then Iₙ = [xⁿ eˣ]₀¹ – ∫₀¹ n xⁿ⁻¹ eˣ dx = e – n Iₙ₋₁. Applying: I₀ = ∫₀¹ eˣ dx = e – 1; I₁ = e – 1·I₀ = e – (e – 1) = 1; I₂ = e – 2·1 = e – 2; I₃ = e – 3(e – 2) = 6 – 2e. Marks are awarded for the clear statement of parts and correct recursive evaluation.
分部积分：设 u = xⁿ, dv = eˣ dx → du = n xⁿ⁻¹ dx, v = eˣ。则 Iₙ = [xⁿ eˣ]₀¹ – ∫₀¹ n xⁿ⁻¹ eˣ dx = e – n Iₙ₋₁。递推求值：I₀ = e – 1; I₁ = e – 1·I₀ = 1; I₂ = e – 2; I₃ = 6 – 2e。评分要求清晰写出分部过程并正确递推。
7. Second Order Differential Equations | 二阶微分方程
Linear second-order ODEs of the form a d²y/dx² + b dy/dx + c y = f(x) appear frequently. The June 2019 paper tested the complementary function (CF) by solving the auxiliary equation am² + bm + c = 0, and the particular integral (PI) using trial functions for polynomial, exponential, or trigonometric f(x). The mark scheme insists on a fully general solution y = CF + PI, and awarding method marks for finding the PI by undetermined coefficients, even if a sign error leads to wrong constants.
形如 a d²y/dx² + b dy/dx + c y = f(x) 的线性二阶常微分方程经常出现。2019年6月试题考查通过求解辅助方程 am² + bm + c = 0 获得补函数 (CF)，以及使用试探函数求特解 (PI) 处理多项式、指数或三角函数形式的 f(x)。评分标准要求最终写出通解 y = CF + PI，并给通过待定系数法求特解的过程赋方法分，即使符号错误导致常数有误。
Example: Solve d²y/dx² – 4 dy/dx + 4y = e²ˣ.
示例： 求解 d²y/dx² – 4 dy/dx + 4y = e²ˣ。
Auxiliary: m² – 4m + 4 = 0 → (m–2)² = 0, repeated root m=2. CF: y = (A + Bx)e²ˣ. For PI, since e²ˣ and xe²ˣ appear in CF, try y = C x² e²ˣ. Differentiate twice and substitute to find C = 1/2. General solution: y = (A + Bx + ½ x²)e²ˣ. Marks are often lost when students fail to multiply by x² despite the repeated root overlap.
辅助方程 m² – 4m + 4 = 0 得重根 m=2，CF 为 y = (A + Bx)e²ˣ。求特解时由于 e²ˣ 和 xe²ˣ 均已在 CF 中，尝试 y = C x² e²ˣ。求导代入定出 C = 1/2。通解为 y = (A + Bx + ½ x²)e²ˣ。考生常因忽略重根导致特解形式不乘以 x² 而失分。
8. Series Expansions – Maclaurin Series | 级数展开 – 麦克劳林级数
Maclaurin series expands a function about x = 0: f(x) = f(0) + f‘(0)x + f’‘(0)x²/2! + f’’‘(0)x³/3! + … . In the June 2019 mark scheme, questions often ask for the series up to x³ or x⁴ for functions like eˣ sin x, ln(1+ sin x), or (1+x)⁻¹². The key is to differentiate repeatedly, evaluate at zero, and be meticulous with algebraic signs. Another common task is to combine known standard series (e.g., eˣ = 1 + x + x²/2! + … , sin x = x – x³/3! + … ) by multiplication or substitution.
麦克劳林级数将函数在 x=0 处展开：f(x) = f(0) + f‘(0)x + f’‘(0)x²/2! + f’’‘(0)x³/3! + … 。2019年6月评分标准中常要求将 eˣ sin x、ln(1+ sin x) 或 (1+x)⁻¹² 等函数展开到 x³ 或 x⁴。核心在于反复求导、代入零并仔细处理符号。另一种常见题型是将已知标准级数（如 eˣ = 1 + x + x²/2! + … , sin x = x – x³/3! + …）通过乘法或代入进行组合。
Example: Find the Maclaurin series for eˣ cos x up to the term in x⁴.
示例： 求 eˣ cos x 的麦克劳林级数，直到 x⁴ 项。
Method 1: Compute derivatives. f(x)=eˣ cos x; f(0)=1; f‘(x)=eˣ cos x – eˣ sin x, f’(0)=1; f’‘(x)= –2eˣ sin x, f’’(0)=0; f‘’‘(x)= –2eˣ (sin x + cos x), f’’‘(0)= –2; f⁽⁴⁾(x)= –4eˣ cos x, f⁽⁴⁾(0)= –4. Series: 1 + x + 0·x²/2 + (–2)x³/6 + (–4)x⁴/24 = 1 + x – x³/3 – x⁴/6. Method 2: Multiply series of eˣ and cos x. The mark scheme accepts both methods, but the multiplication method must be clearly aligned.
方法一：反复求导。f(0)=1, f‘(0)=1, f’‘(0)=0, f’’‘(0)= –2, f⁽⁴⁾(0)= –4。级数为 1 + x – x³/3 – x⁴/6。方法二：将 eˣ 和 cos x 的级数相乘。评分标准两种方法均可，但乘法需严格对齐幂次。
9. Further Vectors – Planes and Distances | 进阶向量 – 平面与距离
Questions on vectors typically involve equations of planes in scalar product form r·n = p, finding the point of intersection of a line and a plane, and calculating the shortest distance from a point to a plane. The June 2019 mark scheme stresses the correct substitution of line equation r = a + λb into the plane equation to solve for λ. The distance formula d = |(AP · n)| / |n| is a regular feature; often students forget the absolute value or mis-calculate the normal vector n from the plane’s equation.
向量题通常涉及平面方程 r·n = p 的点法式、求直线与平面的交点，以及计算点到平面的最短距离。2019年6月评分标准强调将直线方程 r = a + λb 正确代入平面方程求解 λ。距离公式 d = |(AP · n)| / |n| 是常见考点，学生常忘记绝对值或从平面方程中错误计算法向量 n。
Example: Find the perpendicular distance from point P(2, –1, 3) to the plane 2x – y + 2z = 5.
示例： 求点 P(2, –1, 3) 到平面 2x – y + 2z = 5 的垂直距离。
The normal vector n = (2, –1, 2), |n| = √(4+1+4)=3. A point on the plane, e.g., A(0, –5, 0) satisfies 2·0 – (–5) + 0 = 5. Vector AP = (2, 4, 3). Then distance = |(AP·n)| / 3 = |(4 –4 + 6)| / 3 = 6/3 = 2. The mark scheme requires showing the selection of point A and the scalar product steps.
法向量 n = (2, –1, 2)，模为 3。取平面上一点 A(0, –5, 0)，向量 AP = (2, 4, 3)。距离 = |(AP·n)| / 3 = |(4 –4 + 6)| / 3 = 2。评分标准要求展示选取点 A 并进行数量积计算的过程。
10. Inequalities and Numerical Methods | 不等式与数值方法
Further inequality problems may involve proving that a function is always positive or negative by considering its derivative or stationary points. The June 2019 paper included an inequality requiring the use of the Maclaurin expansion to bound an integral, or solving an inequality like eˣ > 1 + x + x²/2 for x > 0. Additionally, numerical methods such as the Newton-Raphson iteration xₙ₊₁ = xₙ – f(xₙ)/f'(xₙ) are tested, with mark scheme emphasis on correct differentiation and an accurate first iteration. For Simpson’s rule, the formula with h/3 must be applied carefully, showing ordinates and multipliers.
进阶不等式问题可能要求通过导数或驻点证明函数恒正或恒负。2019年6月试卷中包含利用麦克劳林展开估计积分值的不等式，或求解类似 eˣ > 1 + x + x²/2（x>0）的问题。此外，数值方法如牛顿–拉弗森迭代 xₙ₊₁ = xₙ – f(xₙ)/f‘(xₙ) 也是考点，评分标准强调正确求导和精确的第一次迭代。辛普森法则需严格套用 h/3 公式，列出纵坐标和乘数。
Example (Newton-Raphson): Use the iteration to find the root of x³ – 2x – 5 = 0 near x=2, performing two iterations.
示例（牛顿迭代）： 对方程 x³ – 2x – 5 = 0 在 x=2 附近的根使用牛顿法，进行两次迭代。
f(x)=x³–2x–5, f‘(x)=3x²–2. x₀=2, f(2)= –1, f’(2)=10, so x₁ = 2 – (–1)/10 = 2.1. Then f(2.1)=0.061, f‘(2.1)=11.23, x₂ = 2.1 – 0.061/11.23 ≈ 2.09457. The mark scheme usually awards marks for the formula, correct substitution, and both iterations.
f(x)=x³–2x–5, f’(x)=3x²–2。x₀=2 时 f(2)= –1，f‘(2)=10，x₁ = 2 – (–1)/10 = 2.1。f(2.1)=0.061，f’(2.1)=11.23，x₂ ≈ 2.09457。评分标准给分点涵盖公式应用、正确代入以及两次迭代计算。
11. Further Integration Techniques | 进阶积分技巧
Integration using partial fractions and inverse trigonometric/hyperbolic substitutions is tested. For example, ∫ dx/(x² – a²) yields (1/2a) ln| (x–a)/(x+a) | + C, while ∫ dx/√(a² – x²) gives arcsin(x/a) + C. The June 2019 mark scheme also expects completion of the square for quadratic denominators, leading to arctan or arsinh forms. Students should remember that ∫ dx/√(x² + a²) = arsinh(x/a) + C = ln(x + √(x²+a²)) + C, and be able to produce logarithmic forms efficiently.
考试中还涉及部分分式法和反三角/双曲代换积分。例如 ∫ dx/(x² – a²) = (1/2a) ln| (x–a)/(x+a) | + C，∫ dx/√(a² – x²) = arcsin(x/a) + C。2019年6月评分标准要求对二次分母配方，进而转为 arctan 或 arsinh 形式。考生应牢记 ∫ dx/√(x² + a²) = arsinh(x/a) + C = ln(x + √(x²+a²)) + C，并能熟练转换为对数形式。
Example: Evaluate ∫ dx/(x² + 2x + 5).
示例： 计算 ∫ dx/(x² + 2x + 5)。
Complete the square: x²+2x+5 = (x+1)² + 4. Then ∫ dx/[(x+1)² + 2²] = (1/2) arctan((x+1)/2) + C.
Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导